Quadratic Equations  Hemant Malhotra

hemant_malhotra
Director at ElitesGrid  CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
Find maximum number of real roots(+ve ,ve) for the equations x^3+6x^2+11x6=0
The max number of +ve real roots of a polynomial f(x)=0 is d number of changes of signs .
the max number of ve real roots of f(x)=0 is number of changes of sign in f(x)=0
so f(x)=x^3+6x^2+11x6
positive real roots = + + +  so 1 sign change so max number of +ve real roots=1
for negative real roots f(x)=x^3+6x^211x6
sign changes are as  +   : 2 changes so 2 negative real roots maxthis rule is known as Descarte's rule of sign. This will give u only max number of real roots .. there is possibility that roots are less than that
Find equation whose roots are reciprocal of the roots of ax^2+bx+c=0
a) cx^2+bx+a
b) bx^2+cx+A=0
c) cx^2+ax+b=0
d) bx^2+ax+c=0Let we have an equation ax^2+bx+C=0and roots of this is p and q now find equation whose roots are reciprocal to roots of this given equation
 reciprocal so just replace x by 1/x
a(1/x)^2+b(1/x)+c=0 so cx^2+bx+a=0
let in same question we want equation whose roots are cube of give equation so
so just replace x by x^1/3
NOTE_ in square we will replace x by x^1/2 , cube by x^1/3 , in cube root x by x^3 and so on :) ... let we have to find equation whose roots are 2 more than the roots of given equation then replace x by x2 .. so we have to follow just reverse process.If sum of two roots of equation x^3px^2+qxr is zero. then
a) pq = r
b) qr = p
c) pr = q
d) pqr = 1sum of roots=a+b+c=p
given a+b=0
so c=p
we know that c is a root of equation so it will satisfy
so p^3p^3+qpr=0
so pq=rIf a and b( does not equal to zero) are roots of x^2 + ax + b = 0 find least value of x^2 + ax + b ( x is real)
a) 9/4
b) 9/4
c) 1/4
d) 1/4a and b are roots of x^2+ax+b
so a+b = a
so 2a+b=0
now product of roots=a*b=b
b(a1)=0case1 a1=0 so a=1
2a+b=0 so b=2case2
b=0but this is given that b is not zero
so a=1 and b=2
so x^2+ax+b ( min)
= D/4a=4ba^2/4=9/4min value of ax^2+bx+c is D/4a
If the equation 2x^2+kx5=0 and x^23x4=0 have one root in common then k
a) 3
b) 27/4
c) 27/4
d) notlet p is common root
2p^2+kp5=0
p^23p4=0
now solve these equations
p^2/(4k15)=p/(5+8)=1/(6k)
so p^2=(4k+15)/(k+6)
and p =3/(k+6)
so (3/k+6)^2=4k+15/(k+6)
so k=3 or 27/4The number of values of pair (a,b) for which a(x+1)^2+b(x^23x2)+x+1=0 is an identity in x
a) 0
b) 1
c) 2
d) infiniteIF ANY QUADRATIC EQUATION ax^2+bx+c=0 HAS MORE THAN TWO ROOTS THEN IT BECOMES AN IDENTITY SO IN THAT CASE a=b=c=0
Number of common roots between two equations x^3+3x^2+4x+5=0 and x^3+2x^2+7x+3=0
a) 0
b) 1
c) 2
d) 3f(x)=x^3+3x^2+4x+5
g(x)=x^3+2x^2+7x+3
f(x)g(x)
x^23x+2=0
x=1,2
( now when u got this root there may be the case that ye value initial equatuion ko satisfy na kare so we should always check this_ these roots are know as false roots ...
put x=1 and x=2 in initial equation alsox^36x^2+15x+3=0 has
a) only one positive root
b) 2 +ve and 1 ve
c) no +ve root
d) notRoll's theroem
it states that if two value a and b are such that f(a)>0 and
f(b) < 0
then there must be a real root in between a and b
Example 
if f(x)=f(x)=2^xx^2+1
f(3)=0
f(3.2) < 0
f(4)=2^416+1>0
so one root will lie between 3 and 4
now f(1)=1/21+1>0
f(2)=1/44+1
so one root lie between 1 and 2
and f(3)=0 so one root 3
so 3 rootsevery equation of an odd degree has at least one real root
x^36x^2+15x+3=0 has atleast one real root and max real roots=3
now question
f(x)=x^36x^2+15x+3
f'(x) ((differentiation)
3x^212x+15=3((( x2)^2+1))
so f'(x)>0 means function is increasing so it will cut x axis only once so one real root only
so this equation has one real root either positive or negative
now f(0)=3
f(1) < 0
so one root will lie between o and 1 so negative real rootx^3ax^2+bxa=0 has three real roots then which is true
a) a=11
b) a not equal to 1
c) b=1
d) b not equal to 1x^3ax^2+bxa=0
(xp)(xq)(xr)=x^3(p+q+r)x^2+(pq+qr+rx)xpqr
now compare
b=pq+qr+rp
a=p+q+r=pqr
(p+q+r/pqr)=1
sk 1/pq+1/qr+1/rp=1
so pq+qr+rp>3 so b does not equal to 1root(x+1)root(x1)=root(4x1) find number of real values of x
a) 1
b) 0
c) 2
d) notwhen you solve, you will get x=5/4 but that value will not satisfy initial equation so zero roots
Find integral value of a for which x^2  2(4a  1)x + 15a^2  2a  7 > 0 is valid for any x
a) 2
b) 3
c) 4
d) noif f(x) > 0 for all values of x
then D < 0
4(4a1)^24(15a^22k7) < 0
a^26a+8 < 0
2 < a < 4
so integral value is 3If roots p,q,r in HP of equation x^3  3ax^2 + 3bx  c = 0 then
a) q=1/p
b) q = b
c) q = c/b
d) q = b/capproach 1
p+q+r = 3a
pq + qr + rp = 3b
pqr = c
q = 2pr/p+r
1 = 2c/q / [3bc/q]
3b = 3c/q
q=c/bapproach 2
roots are in HP
so 1/p,1/q,1/r are the roots of
(1/x)^33z(1/x)^2+3b(1/x)c=0
so cx^3+3bx^23ax+1=0 are in AP
so 1/p+1/q+1/r=3b./c
so 3/q=3b/c
so q=c/bFor all real values of find min and max value of (x^23x+4)/(x^2+3x+4)
x^23x+4/(x^2+3x+4) = y
now make a quadratic in x
x^2(y1)+3x(y+1)+4(y1) = 0
now real x so D > = 0
9(y+1)^216(y1)^2 > = 0
so (7y1)(y7) < = 0
so 1/7 < = y < = 7Number of possible value of integer p for which x^2 + px + 16 = 0 has integral roots
a) 4
b) 6
c) 2
d) notFor roots to be integral Discriminant should be perfect square
D=k^2
a^264=k^2
a^2k^2=64
Now easy ? Answer is 6